anonymous
  • anonymous
Does the following series converge or diverge?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\sum_{n=2}^{infinity} 1/n \sqrt{\ln(n)}\]
anonymous
  • anonymous
I'm wanting to say it converges to 1, but I'm not sure how to prove that. Real analysis wasn't my strong point.
anonymous
  • anonymous
do you know what test to use?

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anonymous
  • anonymous
my original thought was ratio, but i dont know how to apply that here
anonymous
  • anonymous
I'm wanting to say it has something to do with the Cauchy Criterion
anonymous
  • anonymous
...I have no idea what that means..haha
anonymous
  • anonymous
I don't think we covered that
anonymous
  • anonymous
Oh, the ratio test should work.
anonymous
  • anonymous
really? Let me try that really quickly
anonymous
  • anonymous
Well, that gave me infinity over infinity...maybe i'm doing it wrong?
anonymous
  • anonymous
Looking up properties of the natural log right now, give me a sec.
anonymous
  • anonymous
thanks so much!
anonymous
  • anonymous
Integral test would be easier since i see ln(n) and 1/n there.
anonymous
  • anonymous
oh that makes sense!
anonymous
  • anonymous
Ok all i can think of is that \[n / (n+1) \le 1\] and \[\sqrt(\ln(n+1)) \le n\] and \[\sqrt(\ln(n)) \le n\] thus \[\left| n/(n+1) * \sqrt(\ln(n+1)) / \sqrt(\ln(n)) \right| \le \left| 1 * n/n \right| \le 1\]
anonymous
  • anonymous
hmm. Ok, let me think about that for a min :)
anonymous
  • anonymous
hmm actually that might not work cause the ration needs to be less than 1 for the ratio test to work
anonymous
  • anonymous
when i worked it doing the integral test i got infinity meaning divergent does that seem right?

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